Measurements and Theory of Errors - Exam Answer Templates
In the PRC Civil Engineer Licensure Examination, knowing the correct answer is only half the battle — writing it in the expected format is what earns full marks. Examiners use structured rubrics, awarding partial credit for specific terms, formulas, and logical steps. These templates show you exactly how a model examinee writes answers for 1-mark up to long-answer (5-mark) questions on Measurements and Theory of Errors. Study the scoring breakdowns and key phrases carefully: reproducing them in your own answers is the fastest route to a higher board score.
Answer templates
Define 'most probable value' (MPV) in surveying measurements. [1 mark]
Examiner tip
Examiners reward the formula x̄ = Σx/n alongside the word 'arithmetic mean.' A definition in words alone, without the formula, often gets only partial credit in numerical-subject boards.
Model answer
The most probable value (MPV) is the arithmetic mean of a series of repeated measurements of the same quantity under the same conditions. It is the value that has the highest probability of being closest to the true value and is computed as x̄ = Σx / n.
Answer structure
- Line 1: State the definition using the term 'arithmetic mean' and 'repeated measurements' [1 mark]
Scoring breakdown
Criteria
Correct identification that MPV equals the arithmetic mean (x̄ = Σx/n) of repeated observations of the same quantity
Common mark deductions
- Writing 'average value' without linking it to the formula or the concept of probability (-0 to -1 depending on strictness)
- Omitting that measurements must be of the same quantity under the same conditions
- Confusing MPV with the true value (they are not the same)
Key phrases to include
- arithmetic mean
- repeated measurements
- x̄ = Σx / n
- highest probability
- same conditions
Distinguish between systematic errors and random errors in surveying. [2 marks]
Examiner tip
Frame your answer as a contrast — examiners for distinction/comparison questions look for the parallel structure: systematic (law-governed, correctable) vs. random (probabilistic, not individually correctable).
Model answer
Systematic errors follow a definite physical law or cause, always biasing the measurement in the same direction (either consistently positive or consistently negative). They can be corrected by applying computed corrections (e.g., temperature correction to a steel tape). Random (accidental) errors, in contrast, are small residual errors that remain after all systematic corrections are applied; they are equally likely to be positive or negative and are treated statistically using the theory of probability.
Answer structure
- Part A: Define systematic error — follows a law, consistent direction, correctable [1 mark]
- Part B: Define random error — residual, bidirectional, treated statistically [1 mark]
Scoring breakdown
Criteria
Correct description of systematic error: follows a physical law, always in the same direction, and correctable
Criteria
Correct description of random error: small, residual after corrections, equally likely positive or negative, treated statistically
Common mark deductions
- Mentioning only one type and ignoring the other
- Calling random errors 'mistakes' or 'blunders' (those are a third distinct category)
- Failing to state that systematic errors are correctable while random errors are not individually correctable
Key phrases to include
- definite physical law
- same direction (consistently biased)
- correctable
- residual errors
- equally likely positive or negative
- theory of probability
A steel tape has a coefficient of thermal expansion α = 11.6 × 10⁻⁶/°C. A line is measured as 300 m at a field temperature of 40°C. The standard temperature is 20°C. Compute the temperature correction. [2 marks]
Examiner tip
The sign of C_t is conceptually important: if T > T_s, the steel tape expands and is physically longer than its nominal length, so measured distances read short and the correction is positive (add to get true length).
Model answer
Given:
L = 300 m, α = 11.6 × 10⁻⁶/°C, T = 40°C, T_s = 20°C
Formula:
C_t = α L (T − T_s)
Solution:
C_t = (11.6 × 10⁻⁶)(300)(40 − 20)
C_t = (11.6 × 10⁻⁶)(300)(20)
C_t = +0.0696 m
The temperature correction is +0.0696 m (positive because field temperature exceeds standard; the tape is longer, so measured distances read short — the true length is longer).
Answer structure
- Step 1: List all given values with symbols and units [0.5 mark]
- Step 2: Write the formula C_t = αL(T − T_s) [0.5 mark]
- Step 3: Substitute and compute the numerical answer [0.5 mark]
- Step 4: State the sign and physical interpretation [0.5 mark]
Scoring breakdown
Criteria
Correct formula C_t = αL(T − T_s) properly cited
Criteria
Correct substitution and final answer of +0.0696 m with proper sign and unit
Common mark deductions
- Reversing (T_s − T) instead of (T − T_s), giving wrong sign
- Omitting units on the final answer
- Not stating whether the correction is added or subtracted
Key phrases to include
- C_t = αL(T − T_s)
- α = 11.6 × 10⁻⁶/°C
- +0.0696 m
- field temperature exceeds standard
- tape is longer
Three measurements of a horizontal distance are recorded as 100.12 m, 100.08 m, and 100.10 m. Determine (a) the most probable value and (b) the probable error of a single measurement. [3 marks]
Examiner tip
Always write out the residual table in full columnar form (x | v | v²). Board-exam rubrics allocate a dedicated mark for this table. A correct final answer without the table may still lose 1 mark.
Model answer
Given measurements: x₁ = 100.12 m, x₂ = 100.08 m, x₃ = 100.10 m; n = 3
(a) Most Probable Value:
x̄ = (100.12 + 100.08 + 100.10) / 3
x̄ = 300.30 / 3 = 100.10 m
(b) Probable Error of a Single Measurement:
Residuals: v₁ = 100.12 − 100.10 = +0.02 m
v₂ = 100.08 − 100.10 = −0.02 m
v₃ = 100.10 − 100.10 = 0.00 m
Σv² = (0.02)² + (0.02)² + (0.00)² = 0.0004 + 0.0004 + 0 = 0.0008 m²
E = 0.6745 √(Σv² / (n − 1))
E = 0.6745 √(0.0008 / 2)
E = 0.6745 √(0.0004)
E = 0.6745 × 0.02
E = ±0.0135 m
Final Answers:
MPV = 100.10 m
Probable error (single observation) = ±0.0135 m
Answer structure
- Step 1: Compute the arithmetic mean x̄ [1 mark]
- Step 2: Tabulate residuals v = x − x̄ and compute Σv² [1 mark]
- Step 3: Apply E = 0.6745√(Σv²/(n−1)) and compute final answer with sign [1 mark]
Scoring breakdown
Criteria
Correct MPV = 100.10 m
Criteria
Correct residual table with Σv² = 0.0008 m²
Criteria
Correct application of the probable error formula yielding E = ±0.0135 m
Common mark deductions
- Dividing by n instead of n−1 in the formula (uses population standard deviation instead of sample)
- Forgetting the ± sign on the probable error
- Not showing the residual table — step marks require the working to be visible
- Using E = 0.6745√(Σv²/n) — the n−1 denominator is mandatory for sample observations
Key phrases to include
- x̄ = Σx / n
- residuals v = x − x̄
- Σv²
- E = 0.6745√(Σv²/(n−1))
- n − 1 (degrees of freedom)
- ±0.0135 m
Define 'blunder' (mistake) in surveying and explain why it is not treated statistically. [1 mark]
Examiner tip
The key distinguishing phrase is 'gross error' and 'must be eliminated.' Examiners will not award the mark if you say blunders are reduced by taking the mean.
Model answer
A blunder (mistake) is a gross human error caused by carelessness, misreading, or poor judgment (e.g., misreading a rod, recording a wrong digit). Blunders are not treated statistically because they do not follow any probability law — they must be detected and eliminated through field checks, repetition, and verification before statistical analysis begins.
Answer structure
- Line 1: Define blunder as a gross human error with at least one example [0.5 mark]
- Line 2: Explain that it does not follow a probability law and must be eliminated [0.5 mark]
Scoring breakdown
Criteria
Correct definition (gross human error, not following probability law) with statement that blunders must be eliminated, not statistically corrected
Common mark deductions
- Confusing blunders with systematic errors
- Saying blunders are 'minimised' statistically — they are eliminated, not minimised
Key phrases to include
- gross human error
- does not follow probability law
- must be eliminated
- carelessness or misreading
A slope distance of 250 m was measured between two points with an elevation difference of 10 m. Compute the horizontal distance using (a) the exact formula and (b) the approximate slope correction formula. [3 marks]
Examiner tip
State explicitly that slope correction C_h is always negative — this is a high-frequency board-exam pitfall and examiners specifically look for this statement as an indicator of conceptual understanding.
Model answer
Given: L = 250 m (slope distance), h = 10 m (elevation difference)
(a) Exact Formula:
H = √(L² − h²)
H = √(250² − 10²)
H = √(62500 − 100)
H = √62400
H = 249.80 m
(b) Approximate Slope Correction:
C_h = −h² / (2L)
C_h = −(10²) / (2 × 250)
C_h = −100 / 500
C_h = −0.20 m
H ≈ L + C_h = 250 + (−0.20) = 249.80 m
Both methods give H = 249.80 m (the approximate formula is accurate for gentle slopes where h/L < 0.10).
Answer structure
- Step 1: State given data (L and h) [0.5 mark]
- Step 2: Apply exact formula H = √(L² − h²) and compute [1 mark]
- Step 3: Apply approximate formula C_h = −h²/2L and compute H [1 mark]
- Step 4: State that slope correction is always negative (horizontal < slope distance) [0.5 mark]
Scoring breakdown
Criteria
Correct setup and answer using exact formula: H = 249.80 m
Criteria
Correct setup and answer using approximate formula: C_h = −0.20 m, H = 249.80 m
Criteria
Correct sign convention stated (C_h is always negative; horizontal distance is always less than slope distance)
Common mark deductions
- Using h²/2L without the negative sign
- Adding C_h instead of subtracting (wrong sign application)
- Not showing the formula before substituting
- Rounding intermediate values leading to a different final answer
Key phrases to include
- H = √(L² − h²)
- C_h = −h²/2L
- slope correction is always negative
- horizontal distance is always less than slope distance
- 249.80 m
A 30 m tape is found to be actually 30.02 m long under standard conditions. A distance is measured as 450 m using this tape. Find the correct (true) distance. [2 marks]
Examiner tip
Always state the physical logic first: 'tape too long → measured reads short → true is longer.' This guards against formula inversion, which is the most common error on this type of board question.
Model answer
Given:
Nominal tape length = 30 m
Actual tape length = 30.02 m
Measured distance = 450 m
Since the tape is too long (actual > nominal), each 30 m interval laid down by the tape actually covers 30.02 m on the ground — the true distance is greater than the measured reading.
Formula:
True distance = Measured distance × (Actual length / Nominal length)
Solution:
True distance = 450 × (30.02 / 30)
True distance = 450 × 1.000667
True distance = 450.30 m
The correct distance is 450.30 m.
Answer structure
- Step 1: Identify whether the tape is too long or too short and state the direction of error [0.5 mark]
- Step 2: Write the formula: True = Measured × (Actual/Nominal) [0.5 mark]
- Step 3: Substitute values and compute [0.5 mark]
- Step 4: State the final answer with units [0.5 mark]
Scoring breakdown
Criteria
Correct formula: True distance = Measured × (Actual/Nominal) with correct identification that tape too long → true > measured
Criteria
Correct computation and final answer of 450.30 m
Common mark deductions
- Inverting the ratio: using (nominal/actual) instead of (actual/nominal)
- Incorrectly concluding that tape too long → measured reads long (it reads SHORT — each interval is longer, so fewer intervals cover the same ground)
- Not explaining the physical reasoning before the formula
Key phrases to include
- tape is too long (actual > nominal)
- true distance = measured × (actual/nominal)
- true distance is greater than measured distance
- 450.30 m
What is the probable error of the mean for the dataset in T4 (three measurements with E = ±0.0135 m)? [1 mark]
Examiner tip
Relate the formula conceptually: increasing n reduces E_m, demonstrating why surveyors repeat measurements. This statement, if added, shows depth of understanding and can earn bonus credit from lenient examiners.
Model answer
Given: E = ±0.0135 m (probable error of a single observation), n = 3
E_m = E / √n
E_m = 0.0135 / √3
E_m = 0.0135 / 1.732
E_m = ±0.0078 m
The probable error of the mean is ±0.0078 m.
Answer structure
- Line 1: Write formula E_m = E/√n [0.5 mark]
- Line 2: Substitute and compute final answer with sign and units [0.5 mark]
Scoring breakdown
Criteria
Correct formula E_m = E/√n and correct answer of ±0.0078 m
Common mark deductions
- Multiplying by √n instead of dividing
- Forgetting the ± sign
- Omitting units
Key phrases to include
- E_m = E / √n
- ±0.0078 m
- more observations → smaller error of the mean
A surveying party measures four sides of a closed traverse. The probable errors of the individual sides are ±0.015 m, ±0.020 m, ±0.012 m, and ±0.018 m. Compute the probable error of the total perimeter. [2 marks]
Examiner tip
The board exam frequently tests whether a student knows that errors of a sum combine as √(ΣEᵢ²) — NOT as a simple arithmetic sum. Write this formula explicitly before computing to earn the formula mark even if arithmetic is off.
Model answer
Given: E₁ = ±0.015 m, E₂ = ±0.020 m, E₃ = ±0.012 m, E₄ = ±0.018 m
For the sum of independently measured quantities, errors propagate as:
E_total = √(E₁² + E₂² + E₃² + E₄²)
E_total = √[(0.015)² + (0.020)² + (0.012)² + (0.018)²]
E_total = √[0.000225 + 0.000400 + 0.000144 + 0.000324]
E_total = √0.001093
E_total = ±0.0331 m
The probable error of the total perimeter is ±0.0331 m.
Answer structure
- Step 1: State the error propagation formula for sums: E = √(ΣEᵢ²) [1 mark]
- Step 2: Compute each Eᵢ², sum them, take the square root [0.5 mark]
- Step 3: State the final answer with sign and units [0.5 mark]
Scoring breakdown
Criteria
Correct error propagation formula: E_total = √(E₁² + E₂² + E₃² + E₄²)
Criteria
Correct computation yielding E_total = ±0.0331 m
Common mark deductions
- Adding errors directly (E₁ + E₂ + E₃ + E₄) instead of combining in quadrature
- Forgetting to square the individual errors before summing
- Not including the ± sign on the final answer
Key phrases to include
- E_total = √(ΣEᵢ²)
- independent measurements
- errors propagate as square root of sum of squares
- ±0.0331 m
A 30 m tape weighs 0.80 kg (total weight W). It is supported only at its ends and used at a pull P = 80 N. Compute the sag correction per tape length. [3 marks]
Examiner tip
The sag correction is ALWAYS negative — write this statement explicitly. Examiners consistently note that students who drop the negative sign on C_sag lose the interpretation mark even when arithmetic is correct.
Model answer
Given:
W = 0.80 kg → W = 0.80 × 9.81 = 7.848 N (total weight of tape)
L = 30 m (length of tape)
P = 80 N (applied pull)
Formula (using total weight W):
C_sag = −W²L / (24P²)
Solution:
C_sag = −(7.848)² × 30 / (24 × 80²)
C_sag = −(61.59) × 30 / (24 × 6400)
C_sag = −1847.7 / 153600
C_sag = −0.01203 m
The sag correction per tape length is −0.0120 m (negative because sag always shortens the horizontal distance between tape ends).
Answer structure
- Step 1: Convert W from kg to Newtons using W = mg [0.5 mark]
- Step 2: State the sag correction formula C_sag = −W²L/(24P²) [1 mark]
- Step 3: Substitute values and compute [1 mark]
- Step 4: State the sign and physical meaning (sag always negative) [0.5 mark]
Scoring breakdown
Criteria
Correct conversion of weight to Newtons (7.848 N) and identification of the correct sag formula
Criteria
Correct substitution into C_sag = −W²L/(24P²)
Criteria
Correct final answer of −0.0120 m with negative sign stated and explained
Common mark deductions
- Using W in kg without converting to Newtons
- Omitting the negative sign on C_sag
- Using the per-unit-length formula (−w²L³/24P²) incorrectly when total weight W is given
- Confusing P (pull) with W (weight)
Key phrases to include
- C_sag = −W²L/(24P²)
- W in Newtons (convert from kg)
- sag correction is always negative
- supported at ends only
- −0.0120 m
A 50 m tape has a cross-sectional area of 6.0 mm², modulus of elasticity E = 200 GPa, and standard pull P_s = 50 N. It is used at a pull of P = 75 N. Calculate the tension (pull) correction for a measured distance of 300 m. [3 marks]
Examiner tip
Unit conversion is the No. 1 source of wrong answers for C_p. Always write out 'A = 6.0 mm² = 6.0×10⁻⁶ m²' and 'E = 200 GPa = 200×10⁹ N/m²' explicitly — this also earns the conversion marks.
Model answer
Given:
P = 75 N (applied pull)
P_s = 50 N (standard pull)
L = 300 m (measured distance)
A = 6.0 mm² = 6.0 × 10⁻⁶ m²
E = 200 GPa = 200 × 10⁹ Pa = 200 × 10⁹ N/m²
Formula:
C_p = (P − P_s) × L / (A × E)
Solution:
C_p = (75 − 50) × 300 / (6.0 × 10⁻⁶ × 200 × 10⁹)
C_p = (25 × 300) / (1200)
C_p = 7500 / 1200
C_p = +0.00625 m
Note: A × E = 6.0 × 10⁻⁶ m² × 200 × 10⁹ N/m² = 1.20 × 10⁶ N
The tension correction is +0.00625 m (positive because applied pull > standard pull, elongating the tape).
Answer structure
- Step 1: List all given values with correct unit conversions (A in m², E in N/m²) [1 mark]
- Step 2: Write formula C_p = (P − P_s)L / (AE) [0.5 mark]
- Step 3: Substitute and compute AE, then the full expression [1 mark]
- Step 4: State final answer with sign and physical interpretation [0.5 mark]
Scoring breakdown
Criteria
Correct unit conversions: A = 6.0×10⁻⁶ m² and E = 200×10⁹ N/m²
Criteria
Correct formula C_p = (P−P_s)L/(AE) with proper substitution
Criteria
Correct final answer +0.00625 m with sign and interpretation
Common mark deductions
- Failing to convert A from mm² to m² (off by factor of 10⁶)
- Failing to convert E from GPa to Pa/N/m²
- Using wrong formula (e.g., mixing up tension and temperature formulas)
- Dropping the sign of C_p
Key phrases to include
- C_p = (P − P_s)L / (AE)
- A in m² (convert from mm²)
- E in N/m² (convert from GPa)
- +0.00625 m
- P > P_s → tape elongates → positive correction
A line is measured five times, each with a probable error of ±0.03 m. Find the probable error of the sum of all five measurements. [1 mark]
Examiner tip
Know the two square-root formulas: E_sum = E√n (error grows with more measurements in a series) vs. E_m = E/√n (error of the mean shrinks with more measurements). Mixing these up is a perennial board-exam trap.
Model answer
Given: n = 5 equal measurements, each with E = ±0.03 m
For n measurements of equal probable error E, the probable error of the sum is:
E_sum = E × √n
E_sum = 0.03 × √5
E_sum = 0.03 × 2.236
E_sum = ±0.0671 m
The probable error of the sum is ±0.0671 m.
Answer structure
- Line 1: Write formula E_sum = E√n [0.5 mark]
- Line 2: Substitute and compute final answer [0.5 mark]
Scoring breakdown
Criteria
Correct formula E_sum = E√n applied to n = 5 equal-error measurements, giving ±0.0671 m
Common mark deductions
- Multiplying directly: E × n = 0.15 m (ignoring the square root)
- Confusing this with the error of the mean formula (E/√n)
Key phrases to include
- E_sum = E√n
- equal probable errors
- ±0.0671 m
A 30 m tape is actually 29.97 m long. A baseline is measured as 600 m. Find the true length of the baseline and state the direction of correction. [2 marks]
Examiner tip
The conceptual logic is: a short tape requires more intervals to cover the same distance → the tape reading (number of intervals × nominal length) exceeds the true distance → correction is negative. Memorise this logic, not just the formula.
Model answer
Given:
Nominal length = 30 m
Actual length = 29.97 m (tape is too short)
Measured distance = 600 m
Physical reasoning: The tape is shorter than nominal. Each 30 m reading actually covers only 29.97 m on the ground — the true distance is less than the measured distance.
Formula:
True distance = Measured × (Actual / Nominal)
True distance = 600 × (29.97 / 30)
True distance = 600 × 0.999
True distance = 599.40 m
The correction is −0.60 m (negative; tape too short → subtract from measured distance).
Answer structure
- Step 1: Identify tape is too short and state direction (true < measured) [0.5 mark]
- Step 2: Apply formula True = Measured × (Actual/Nominal) [0.5 mark]
- Step 3: Compute True = 599.40 m [0.5 mark]
- Step 4: State correction = −0.60 m with correct sign and direction [0.5 mark]
Scoring breakdown
Criteria
Correct formula and physical logic: tape too short → true < measured
Criteria
Correct answer of 599.40 m with negative correction of −0.60 m stated
Common mark deductions
- Using nominal/actual instead of actual/nominal
- Concluding that tape too short → measured reads short (wrong — measured reads LONG when tape is short)
- Adding the correction instead of subtracting
Key phrases to include
- tape too short (actual < nominal)
- true distance is less than measured distance
- True = Measured × (Actual/Nominal)
- 599.40 m
- correction = −0.60 m
A surveying team measures a line four times with results 250.12, 250.08, 250.15, and 250.09 m. Determine: (a) the most probable value, (b) the probable error of a single observation, (c) the probable error of the mean, and (d) state what each probable error value physically represents. [5 marks]
Examiner tip
For 5-mark long-answer questions on statistics, the residual table with the Σv = 0 verification is mandatory. Examiners specifically allocate 1 mark for the table setup. The interpretation in part (d) is what separates 4-mark from 5-mark answers — never skip it.
Model answer
Given measurements (n = 4):
x₁ = 250.12 m, x₂ = 250.08 m, x₃ = 250.15 m, x₄ = 250.09 m
(a) Most Probable Value (MPV):
x̄ = (250.12 + 250.08 + 250.15 + 250.09) / 4
x̄ = 1000.44 / 4
x̄ = 250.11 m
(b) Probable Error of a Single Observation:
Residual Table:
│ x (m) │ v = x − x̄ (m) │ v² (m²) │
│ 250.12 │ +0.01 │ 0.0001 │
│ 250.08 │ −0.03 │ 0.0009 │
│ 250.15 │ +0.04 │ 0.0016 │
│ 250.09 │ −0.02 │ 0.0004 │
│ Σ │ 0.00 ✓ │ 0.0030 │
Check: Σv = +0.01 − 0.03 + 0.04 − 0.02 = 0.00 ✓ (verifies x̄)
E = 0.6745 √(Σv² / (n−1))
E = 0.6745 √(0.0030 / 3)
E = 0.6745 √(0.0010)
E = 0.6745 × 0.031623
E = ±0.0213 m
(c) Probable Error of the Mean:
E_m = E / √n
E_m = 0.0213 / √4
E_m = 0.0213 / 2
E_m = ±0.0107 m
(d) Physical Interpretation:
E = ±0.0213 m means there is a 50% probability that any single measurement of this line differs from the true value by no more than 0.0213 m.
E_m = ±0.0107 m means there is a 50% probability that the computed mean (MPV = 250.11 m) differs from the true value by no more than 0.0107 m — the mean is a more reliable estimate of the true distance than any individual measurement.
Final Answers:
MPV = 250.11 m
E (single observation) = ±0.0213 m
E_m (mean) = ±0.0107 m
Answer structure
- Part (a): Compute x̄ = Σx/n [1 mark]
- Part (b): Set up residual table with v = x − x̄ and v²; verify Σv = 0; apply E = 0.6745√(Σv²/(n−1)) [2 marks]
- Part (c): Apply E_m = E/√n [1 mark]
- Part (d): State physical meaning of E and E_m in terms of 50% probability [1 mark]
Scoring breakdown
Criteria
Correct MPV = 250.11 m
Criteria
Correct residual table with Σv² = 0.0030 m² and verification that Σv = 0
Criteria
Correct application of E = 0.6745√(Σv²/(n−1)) yielding E = ±0.0213 m
Criteria
Correct E_m = E/√n = ±0.0107 m
Criteria
Correct interpretation: 50% probability statement for both E and E_m, explaining why E_m < E
Common mark deductions
- Omitting the residual table and jumping straight to Σv² — loses the table mark
- Not verifying Σv = 0 (this is an important self-check that examiners look for)
- Using n instead of n−1 in the denominator
- Omitting the physical interpretation in part (d)
- Leaving out the ± sign on both E and E_m
Key phrases to include
- x̄ = Σx/n
- residual v = x − x̄
- Σv = 0 (verification check)
- E = 0.6745√(Σv²/(n−1))
- E_m = E/√n
- 50% probability
- mean is more reliable than a single observation
- ±0.0213 m
- ±0.0107 m
A tape of nominal length 50 m is standardised at 20°C. The coefficient of thermal expansion is α = 11.6 × 10⁻⁶/°C, cross-sectional area A = 6.45 mm², E = 200 GPa, standard pull P_s = 50 N. In the field, temperature is 35°C, applied pull is 80 N, and the tape is supported throughout (no sag). A distance is measured as 500 m. Find the total corrected distance. [5 marks]
Examiner tip
For multi-correction tape problems, always solve in sequence: C_t → C_p → C_sag → C_total → True distance. The structured format earns step marks even when the final answer is slightly off due to rounding. Examiners follow the same sequence in their rubric.
Model answer
Given:
L (measured) = 500 m, nominal tape = 50 m
α = 11.6 × 10⁻⁶/°C, T = 35°C, T_s = 20°C
A = 6.45 mm² = 6.45 × 10⁻⁶ m²
E_steel = 200 GPa = 200 × 10⁹ N/m²
P_s = 50 N, P = 80 N
No sag (tape fully supported)
Step 1 — Temperature Correction:
C_t = α × L × (T − T_s)
C_t = (11.6 × 10⁻⁶)(500)(35 − 20)
C_t = (11.6 × 10⁻⁶)(500)(15)
C_t = +0.0870 m
(Positive: field temp > standard → tape longer → measured reads short → add)
Step 2 — Tension (Pull) Correction:
C_p = (P − P_s) × L / (A × E)
A × E = 6.45 × 10⁻⁶ × 200 × 10⁹ = 1.29 × 10⁶ N
C_p = (80 − 50) × 500 / (1.29 × 10⁶)
C_p = (30 × 500) / 1290000
C_p = 15000 / 1290000
C_p = +0.01163 m
(Positive: applied pull > standard → tape stretched → measured reads short → add)
Step 3 — Sag Correction:
C_sag = 0 (tape fully supported throughout — no sag)
Step 4 — Total Correction:
C_total = C_t + C_p + C_sag
C_total = +0.0870 + 0.01163 + 0
C_total = +0.09863 m
Step 5 — True (Corrected) Distance:
True distance = Measured + C_total
True distance = 500 + 0.09863
True distance = 500.099 m ≈ 500.10 m
Final Answer: The corrected horizontal distance is 500.10 m.
Answer structure
- Step 1: Temperature correction C_t = αL(T−T_s) with sign [1 mark]
- Step 2: Tension correction C_p = (P−P_s)L/(AE) with unit conversion and sign [1.5 marks]
- Step 3: Sag correction = 0, with reason stated [0.5 mark]
- Step 4: Sum all corrections [0.5 mark]
- Step 5: Add C_total to measured distance for final answer [0.5 mark]
- Correct final answer with units [0.5 mark if not already credited above; combined with Step 5]
Scoring breakdown
Criteria
Correct C_t = +0.0870 m with sign and physical explanation
Criteria
Correct unit conversions for A and E, and correct C_p formula
Criteria
Correct C_p = +0.01163 m
Criteria
Correct recognition that C_sag = 0 because tape is fully supported, and correct summation of corrections
Criteria
Correct final corrected distance of 500.10 m
Common mark deductions
- Forgetting to state that C_sag = 0 when tape is fully supported
- Unit conversion error for A (keeping it in mm²) or E (keeping it in GPa without converting)
- Adding all corrections without checking individual signs
- Not writing the formula for each correction before substituting — formula marks are allocated separately
Key phrases to include
- C_t = αL(T−T_s) = +0.0870 m
- C_p = (P−P_s)L/(AE) = +0.01163 m
- tape fully supported — C_sag = 0
- A in m², E in N/m²
- C_total = +0.09863 m
- True distance = Measured + C_total
- 500.10 m
Mark wise strategy
State the definition, formula, or direct answer immediately. For numerical questions, write Given → Formula → Answer in three compact lines. Do not waste time on lengthy preambles.
Do
- Use exact technical terms (e.g., 'arithmetic mean,' 'systematic error,' 'probable error')
- Include the relevant formula (e.g., x̄ = Σx/n) even for definition questions
- Include units and sign (±) on all numerical answers
- Write legibly — examiners scan quickly for key terms
Don't
- Do not write long paragraphs — one clear sentence is sufficient
- Do not use informal terms or paraphrases in place of standard surveying terminology
- Do not skip units or signs on numerical answers
For comparison/distinction questions, structure as Part A vs. Part B with a clear contrast. For numerical problems, show all steps: Given → Formula → Substitution → Answer. Each logical step is worth one mark.
Do
- Show the formula explicitly before substituting numerical values
- For correction problems, state the sign and the physical reason for the sign
- For statistical problems, show at least the residual expression (v = x − x̄)
- State the final answer as a complete sentence with value, unit, and sign
Don't
- Do not jump directly to the numerical answer without showing the formula
- Do not use the same sign for tape-too-long and tape-too-short corrections
- Do not omit conversion of units (mm² to m², GPa to Pa)
Use a systematic format: (1) List all given data with units, (2) State the formula with citation of the variable names, (3) Show the computation step by step, (4) State the sign and physical interpretation of the result. For statistical questions, include the full residual table.
Do
- Set up the residual table (x | v | v²) for all statistical problems — it is worth a dedicated mark
- Verify Σv = 0 after computing residuals as a self-check
- Explicitly convert all units before substituting into formulas
- Label each part (a), (b), (c) clearly if the question has sub-parts
Don't
- Do not mix up the formulas for E (probable error of single obs) and E_m (probable error of the mean)
- Do not forget n−1 (not n) in the denominator of the probable error formula
- Do not omit the physical interpretation — it is often where the final mark is allocated
Treat each sub-part as a mini 1-mark to 2-mark question. Write in a structured report style: (1) Given data table, (2) Formula section, (3) Computation section, (4) Summary table of corrections (for tape problems), (5) Final answer with interpretation. Never skip parts even if uncertain — partial marks are available.
Do
- For multi-correction tape problems: solve C_t, C_p, C_sag in sequence and tabulate them before summing
- For statistical problems: include the full residual table, verify Σv = 0, compute E, then E_m
- State physical meaning of all final answers (e.g., 'the correction is positive because the tape is longer than nominal')
- Use boxed or underlined final answers for each sub-part to aid examiner navigation
- Include a sketch or diagram for slope and sag problems
Don't
- Do not skip sub-parts (d) or (e) even if earlier parts are uncertain — interpretation marks are independent
- Do not round intermediate answers — carry at least 5 significant figures until the final step
- Do not write in a disorganised stream-of-consciousness style — use clear step numbering
General answer writing tips
- Always start a numerical problem by writing the given data and the formula before substituting values — examiners award a mark for correct formula citation even if arithmetic errors occur later.
- State units explicitly at every step (metres, °C, N, MPa). A correct numerical answer without units will usually lose the final mark.
- For tape correction problems, state the sign convention (positive correction = distance is longer than measured; negative = shorter) before computing. This prevents sign errors and signals conceptual clarity to the examiner.
- When asked to classify errors, use the exact terminology: blunder/mistake, systematic error, random (accidental) error. Paraphrasing without the standard term often forfeits the mark.
- Draw a neat, labelled sketch for slope-distance and sag problems. Even a simple triangle showing L, H, and h earns a diagram mark and helps you set up the formula correctly.
- For probable-error questions, set up the residual table (x, v = x − x̄, v²) in columnar form. Missing the table structure is the single most common reason for losing step marks.
- When a tape-too-long/short scenario is given, explicitly state whether you are measuring or laying out — the correction sign reverses between the two cases, and an unprompted statement shows the examiner you know the distinction.
- Check significant figures: board-exam answers typically expect 4 significant figures for distance corrections and 3 for statistical quantities. Round only at the final step.
Diagrams
Tape Correction Decision Flowchart — Which Formula to Apply
A decision-tree flowchart guiding students through the systematic application of each tape correction formula. Students follow the branches to determine the correct formula and sign for temperature, tension, sag, and slope corrections, then combine them to obtain the true distance.
Measurements and Theory of Errors — Master Concept Map
A comprehensive mind map showing the full hierarchy of topics in Measurements and Theory of Errors: error types, statistical analysis formulas, all tape correction types with their formulas and sign rules, and error propagation rules.
Probable Error Calculation — Step-by-Step Procedure
A step-by-step flowchart for computing the most probable value, probable error of a single observation, and probable error of the mean. The verification step (Σv = 0) is explicitly included as a quality check, matching the procedure expected in board exam answers.
Distribution of Tape Correction Question Types in PRC Board Exams
An approximate distribution of tape correction question types appearing in PRC Civil Engineer board examinations based on historical exam patterns. Tape too long/short and temperature correction together account for over 55% of tape-related questions, making them the highest-priority topics for review.